Question

What is the sum, in degrees, of the measures of the interior angles of a pentagon?

Answer 1

`540^o` or `5xx108^o`

Explanation:

In any `n`-gon, the sum of the interior angles is `(n-2)*180^o`.
This can be proven by taking a regular n-gon (why not?), and drawing a (isosceles) triangle from the centre to one of the sides.

The centre angle will be `360^o/n` and the other angles of that triangle will be `1/2(180-360/n)`.
The inner angle of the n-gon will be just twice that, or `180-360/n`

The sum of all inner angles will then be:
`n*(180-360/n)=n*(180xxn/n-360/n)=`

`canceln*((180n-360)/canceln)=(n-2)*180^o`

In case of a regular n-gon, each angle will be `(n-2)/n*180^o`